Question

The values of two resistors are $R_1=(6\pm 0.3)k\Omega$ and $R_2=(10\pm 0.2)k\Omega$ . The percentage error in the equivalent resistance when they are connected in parallel is

• A. 5.125%
• B. 2%
• C. 3.125%
• D. 7%
• E. 10.125%

Answer 1

Answer: (E)

$R_1=(6\pm 0.3)k\Omega ,R_2=(10\pm 0.2)k\Omega$
$R_{parallel}=\frac{R_1{R}_2}{(R_1+R_2)}$
Let $(R_1+R_2)=x$
$\Rightarrow$ $R_P=\frac{R_1{R}_2}{x}$
$\ln R_p=\ln R_1+\ln R_2-\ln x$
Differentiating, $\frac{\Delta R_p}{R_p}=\frac{\Delta R_1}{R_1}+\frac{\Delta R_2}{R_2}+\left(-\frac{\Delta x}{x}\right)$ $\Delta x_{mean}=\frac{0.3+0.2}{2}=0.25\Omega$
$R_{mean}=\frac{6+10}{2}=8\Omega$
$\therefore$ $x=\frac{6+10}{2}=8\Omega$
$\Rightarrow$ $\frac{\Delta x}{x}=\frac{0.25}{8}$
$\therefore$ Total error $=\frac{0.3}{6}+\frac{0.2}{10}+\frac{0.25}{8}$ $=0.05+0.02+0.03125=0.10125$
$\therefore$ $\frac{\Delta R_p}{R_p}=10.125$ %